Water Fountain Shows Cool Physics

Here is a clever water fountain in Japan.

What did I notice at first? Look at the spaces that make the letters as they fall. They get bigger. Why?

How about we start with a slightly simpler case. Suppose I build a water fountain that just releases two drops of water one after the other. Maybe the the second drop is released from the same point, but 0.2 seconds later. It seems to make sense that the two drops would stay 0.2 seconds apart. And they do.

Ok, to illustrate what is going on I have created a quick vpython simulation. Here you can see what that would look like.

This does appear to have the same effect as the Japanese water fountain. As the two drops fall, the distance between them increases. Here is a plot of the vertical position of the two water drops as a function of time.

Just for fun, let me also plot the separation of the two drops as a function of time.

Except for the short time when the second drop hasn’t yet started to fall, the distance between the drops increases at a constant rate. The longer they fall, the further apart they get.

Does this all make sense? Maybe you are thinking: but if they are dropped 0.2 seconds apart, shouldn’t they hit the bottom 0.2 seconds apart? Yes, and they do. If you look at the data from the simulation, the first water drop hits the bottom at 1.74 seconds. The second drop hits the bottom at 1.94 seconds – a difference of 0.2 seconds. Since both water drops are moving faster, a 0.2 second time difference will mean a larger vertical difference in position.

Let me show this algebraically. If an object is in free fall, it will have a constant acceleration of -9.8 m/s2 in the vertical direction. What is the position of the first drop as a function of time? I could re-derive the kinematic equation, but I will just pull it out for now. If an object has a constant acceleration then the following is true:

Maybe my notation isn’t quite clear. Here, y1 is the vertical position of the first water drop. I am assuming it started to move at time t = 0 seconds. The y10 is the initial vertical position of this first water drop. Yes, it is a little bit confusing. Let me clear things up by saying that the water drop started at a position h and its initial vertical velocity was zero m/s. This means I can re-write that as:

Now for water drop two. It also starts at the same position with the same initial velocity and the same acceleration. However, it does not start at time t = 0 seconds. Instead it starts after some delay. Let me call this time delay td. This would make the position of the second drop look like (after a time td):

Why is it (t – td)? Where should water drop 2 be at time t = td? It should be at h. So, this expression does seem to work. Of course during the time before t = td, this expression doesn’t really work.

Now to get an expression for the separation between the two drops. I will call this s so that:

Some interesting things:

Just like the above plot of the separation, this expression says that it should increase with time. The only variable in this equation is the time (at least for a given set of water drops).

Does this have the correct units? m/s2 times seconds squared does indeed give units of meters.

The slope of this line is gtd. If you were able to find the slope of the above plot of the separation, you would get 1.96 m/s which is indeed the same as (9.8 m/s2)(.2 s).

Doesn’t this expression give a negative separation at t = 0? Yes. However, this expression isn’t even valid until t = td. At that time, the separation is (1/2)g(td)2 which is exactly how far the first drop would fall in that amount of time.

So, the fountain is just simple kinematics. Some see the technology in the fountain. Others see it as art. I see it as physics.